Root locus — Evans' method, rules for sketching

3.5.40 · Physics › Guidance, Navigation & Control (GNC)

WHAT is the root locus?

WHY yeh equation? Closed-loop transfer function hai:

T (s) = \frac{K G ( s )}{1 + K G ( s ) H ( s )} .

Poles wahan hote hain jahan denominator zero hota hai, yaani $1 + K G (s) H (s) = 0$ . Neeche ki saari baatein is ek line se nikaalti hain.

Deriving the two conditions from scratch

$L (s) = K G (s) H (s) = K \frac{N ( s )}{D ( s )}$ likho. Characteristic equation:

1 + L (s) = 0 ⟹ L (s) = - 1.

Ab $- 1$ ek complex number hai jiska magnitude 1 aur angle $180°$ hai (ya koi bhi odd multiple). $L (s) = - 1$ ko magnitude aur angle mein split karo:

The sketching rules (each DERIVED, not memorized)

Maano $n$ = open-loop poles ki sankhya (roots of $D$ ), $m$ = open-loop zeros ki sankhya (roots of $N$ ). Maano $n \geq m$ .

Rule 1 — Start and end. $1 + K \frac{N}{D} = 0$ ko $D (s) + K N (s) = 0$ ke roop mein likho.

$K \to 0$ : $D (s) = 0$ → branches open-loop poles par start hoti hain.
$K \to \infty$ : $K$ se divide karo, $N (s) = 0$ → branches open-loop zeros par end hoti hain. $n$ branches hoti hain; $m$ finite zeros par end hoti hain, baaki $n - m$ infinity tak jaati hain.

Rule 2 — Symmetry. Coefficients real hain, isliye complex poles conjugate pairs mein aate hain → locus real axis ke baare mein symmetric hoti hai.

Rule 3 — Real-axis segments. Ek real point locus par hoga agar uske daayein taraf real poles aur zeros ki sankhya odd ho. WHY? Complex-conjugate pairs ke angle contributions cancel ho jaate hain; daayein taraf ka har real singularity $180°$ contribute karta hai. Odd count → total odd multiple of $180°$ hai → angle condition satisfy hoti hai.

Rule 4 — Asymptotes. $n - m$ escaping branches seedhi lines ki taraf jaati hain jinke angles hain:

θ_{a} = \frac{\pm 180° ( 2 k + 1 )}{n - m}

jo real axis par centroid par milti hain:

σ_{a} = \frac{\sum poles - \sum zeros}{n - m} .

WHY centroid? Bade $∣ s ∣$ ke liye, polynomial division ke pehle do terms match karne ke baad $G (s) H (s) \approx \frac{1}{( s - σ _{a} ) ^{n - m}}$ ho jaata hai — door se dekhne par yeh centroid par multiplicity $n - m$ ka ek single pole jaisa lagta hai.

Rule 5 — Breakaway/break-in points. Jahan branches real axis se nikaalti ya wapas milti hain, do poles coincide karte hain → $K$ local extremum par hota hai. Solve karo:

\frac{d K}{d s} = 0, K = - \frac{D ( s )}{N ( s )} .

WHY? Double root par polynomial mein repeated root hoti hai, isliye $K (s)$ turn karta hai: $d K / d s = 0$ .

Rule 6 — $j ω$ -axis crossing. $D (s) + K N (s)$ par Routh–Hurwitz use karo taaki woh $K$ milein jo poori row zero kare; auxiliary equation se crossing frequency $ω$ milti hai. Yeh stability ke liye critical gain hai.

Rule 7 — Angle of departure/arrival (complex pole/zero $p$ se):

θ_{d} = 180° + \sum ∠ (zeros \to p) - \sum ∠ (other poles \to p) .

WHY? Angle condition ko $p$ ke infinitesimally kareeb ek test point par apply karo.

Figure — Root locus — Evans' method, rules for sketching

Worked Example 1 — $G (s) H (s) = \frac{1}{s ( s + 2 )}$

Poles: $s = 0, s = - 2$ ( $n = 2$ ). Zeros: koi nahi ( $m = 0$ ).

Step — Real axis segment. Why? $0$ aur $- 2$ ke beech, exactly ek pole ( $s = 0$ ) kisi bhi interior point ke daayein hota hai → odd → segment $[- 2, 0]$ locus par hai. ✔

Step — Asymptotes. $n - m = 2$ isliye $θ_{a} = 90°, 270°$ . Centroid $σ_{a} = \frac{( 0 ) + ( - 2 ) - 0}{2} = - 1$ . Why centroid $- 1$ par? Yeh poles ka average hai. Branches $s = - 1$ se seedha upar aur neeche shoot karti hain.

Step — Breakaway. $K = - D / N = - s (s + 2) = - (s^{2} + 2 s)$ . Set $d K / d s = - (2 s + 2) = 0 \Rightarrow s = - 1$ . Why yahan? Symmetry ne demand kiya tha ki breakaway do real poles ke beech ho; algebra ne $s = - 1$ confirm kiya.

Step — Gain at breakaway. $K = - (- 1) (- 1 + 2) = 1$ . Toh $K = 1$ par $- 1$ par double pole milta hai; uske aage poles real part $- 1$ ke saath complex ho jaate hain (constant damping).

Insight: Yeh system sabhi $K > 0$ ke liye stable hai — vertical asymptote kabhi right half plane mein cross nahi karta.

Worked Example 2 — $G (s) H (s) = \frac{1}{s ( s + 1 ) ( s + 2 )}$

Poles $0, - 1, - 2$ ( $n = 3$ ), $m = 0$ .

Asymptotes. $θ_{a} = 60°, 180°, 300°$ ; $σ_{a} = \frac{- 3}{3} = - 1$ . Why 3 asymptotes? $n - m = 3$ branches saari escape karti hain.

$j ω$ crossing (Routh). Char. eqn: $s^{3} + 3 s^{2} + 2 s + K = 0$ . Routh array ke first column ko chahiye $\frac{3 \cdot 2 - K}{3} > 0 \Rightarrow K < 6$ . $K = 6$ par: auxiliary $3 s^{2} + 6 = 0 \Rightarrow s = \pm j 2$ . Why yeh matter karta hai: locus imaginary axis ko $ω = 2$ par cross karti hai jab $K = 6$ — yeh maximum stable gain hai.

Common mistake Steel-manning the classic errors

"Branches infinity par end hoti hain, toh zeros matter nahi karte." Theek lagta hai kyunki bahut saare textbook systems mein finite zeros nahi hote. Fix: Sirf $n - m$ branches infinity jaati hain; ek zero add karna ek branch ko apni taraf kheenchta hai aur locus ko wapas left half plane mein mod sakta hai — zeros tumhara stabilizing tool hai (yahi toh ek PD controller add karta hai).

"Real-axis rule = daayein sirf poles hone chahiye." Theek lagta hai kyunki poles wahan hote hain jahan branches start hoti hain. Fix: daayein poles aur zeros dono aate hain, aur count odd hona chahiye.

"Centroid woh jagah hai jahan poles axis cross karti hain." Theek lagta hai kyunki yeh real axis par hi hoti hai. Fix: $σ_{a}$ sirf asymptote intersection hai; actual crossing Routh se aati hai, aur woh generally alag hoti hain.

Recall Feynman: explain to a 12-year-old

Socho ek table par do magnets (poles) hain aur ek dial hai. Jab dial zero par hota hai, do chhote beads exactly magnets par baithe hain. Jab tum dial upar karte ho, beads fixed paths par slide karte hain — kabhi "target dots" (zeros) ki taraf, kabhi straight ramps par edges ki taraf ud jaate hain. Root locus bas un bead paths ki drawing hai. Agar koi bead ek red danger line (imaginary axis) cross kare, toh machine uncontrollably kaanpne lagti hai (unstable). Tum drawing use karte ho ek dial setting chunne ke liye jo sabhi beads ko safe side par rakhe.

Flashcards

Root locus ke har point ko define karne wali equation kya hai?

Characteristic equation

1 + K G (s) H (s) = 0

, yaani

G (s) H (s) = - 1

Locus ki SHAPE kaun si condition determine karti hai (K se independent)?

Angle condition

∠ G (s) H (s) = \pm 180° (2 k + 1)

Chosen point par gain K kaun si condition deti hai?

Magnitude condition

K = 1/∣ G (s) H (s) ∣

K ke 0→∞ jaane par branches kahan start aur end hoti hain?

Open-loop poles par start (K→0), open-loop zeros ya infinity par end (K→∞).

Kitni branches infinity jaati hain?

n - m

(poles minus zeros).

Locus ke liye real-axis rule kya hai?

Ek point locus par hoga agar uske DAAYEIN real poles aur zeros ki sankhya odd ho.

Asymptote angles formula kya hai?

θ_{a} = \pm 180° (2 k + 1) / (n - m)

Asymptote centroid formula kya hai?

σ_{a} = (\sum poles - \sum zeros) / (n - m)

Breakaway points kaise dhoondhe jaate hain?

d K / d s = 0

solve karo jahan

K = - D (s) / N (s)

Locus imaginary axis kahan cross karti hai yeh kaise pata karte hain?

Critical K dhoondne ke liye Routh–Hurwitz apply karo, phir

ω

ke liye auxiliary equation use karo.

Locus real axis ke baare mein symmetric kyun hoti hai?

Kyunki polynomial ke real coefficients hain, isliye complex poles conjugate pairs mein aate hain.

1/ [s (s + 2)]

ke liye breakaway point aur uska gain kahan hai?

s = - 1

K = 1

Connections

Characteristic equation & closed-loop poles
Routh–Hurwitz stability criterion
PID / PD controller design — locus reshape karne ke liye zeros add karna
Bode plot & frequency response — magnitude/phase ka alternate view
Nyquist criterion — $L (s) = - 1$ ka encirclement view
Damping ratio and settling time — pole locations se $ζ, ω_{n}$ padhna
GNC control loops — autopilots mein gain tuning kahan fit hoti hai